std::ranges::destroy
來自 cppreference.com
定義於標頭檔案 <memory> |
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呼叫簽名 (Call signature) |
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template< no-throw-input-iterator I, no-throw-sentinel-for<I> S > requires std::destructible<std::iter_value_t<I>> |
(1) | (C++20 起) |
template< no-throw-input-range R > requires std::destructible<ranges::range_value_t<R>> |
(2) | (C++20 起) |
1) 銷燬範圍
[
first,
last)
中的物件,如同透過以下方式:for (; first != last; ++first) std::ranges::destroy_at(std::addressof(*first)); return first;
本頁描述的類函式實體是 演算法函式物件(非正式地稱為 niebloids),即
目錄 |
[編輯] 引數
first, last | - | 定義要銷燬元素的範圍的迭代器-哨兵對 |
r | - | 要銷燬的range |
[編輯] 返回值
一個迭代器與 last 比較相等。
[編輯] 複雜度
與 first 和 last 之間的距離呈線性關係。
[編輯] 可能實現
struct destroy_fn { template<no-throw-input-iterator I, no-throw-sentinel-for<I> S> requires std::destructible<std::iter_value_t<I>> constexpr I operator()(I first, S last) const noexcept { for (; first != last; ++first) std::ranges::destroy_at(std::addressof(*first)); return first; } template<no-throw-input-range R> requires std::destructible<std::ranges::range_value_t<R>> constexpr std::ranges::borrowed_iterator_t<R> operator()(R&& r) const noexcept { return operator()(std::ranges::begin(r), std::ranges::end(r)); } }; inline constexpr destroy_fn destroy{}; |
[編輯] 示例
以下示例演示如何使用 ranges::destroy
銷燬連續的元素序列。
執行此程式碼
#include <iostream> #include <memory> #include <new> struct Tracer { int value; ~Tracer() { std::cout << value << " destructed\n"; } }; int main() { alignas(Tracer) unsigned char buffer[sizeof(Tracer) * 8]; for (int i = 0; i < 8; ++i) new(buffer + sizeof(Tracer) * i) Tracer{i}; //manually construct objects auto ptr = std::launder(reinterpret_cast<Tracer*>(buffer)); std::ranges::destroy(ptr, ptr + 8); }
輸出
0 destructed 1 destructed 2 destructed 3 destructed 4 destructed 5 destructed 6 destructed 7 destructed
[編輯] 參閱
(C++20) |
銷燬範圍內的多個物件 (演算法函式物件) |
(C++20) |
銷燬給定地址處的物件 (演算法函式物件) |
(C++17) |
銷燬物件範圍 (函式模板) |